SOLVING LOGARITHMIC EQUATIONS

Note:

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Solve for x in the following equation.

Problem 8.6d:

$\log _{.982}\left( 4x^{2}-7x-10\right) =20$

Answers:T he exact answers are $x=\displaystyle \frac{7\pm \sqrt{ 209+16\cdot \left( .982\right) ^{20}}}{8}.$ The approximate answers are $x\approx 2.729582=\quad$and $\quad -0.979582.$

Solution:

The above equation is valid only if $\quad \log _{.982}\left( 4x^{2}-7x-10\right) =20$ is valid. The term $\log _{.982}\left( 4x^{2}-7x-10\right)$ is valid if $\left( 4x^{2}-7x-10\right) >0\longrightarrow x>\displaystyle \frac{7+\sqrt{209}}{8}\approx 2.6821$ or $x<\displaystyle \frac{7- \sqrt{209}}{8}\approx -0.9321.$ Therefore, the equation is valid when the domain is the set of real numbers less than $$\frac{7-\sqrt{209}}{8}$$ or greater than $$\frac{7+\sqrt{209}}{8}.$$

Covert the logarithmic equation to an exponential equation with base e.

$$\begin{eqnarray*}&& \\ \log _{.982}\left( 4x^{2}-7x-10\right) &=&20 \\ && \\ ... ... 4\right) \left( 10+\left( .982\right) ^{20}\right) }}{8} \\ && \end{eqnarray*}$$

$$\begin{eqnarray*}&& \\ x &=&\displaystyle \frac{7\pm \sqrt{209+16\cdot \left( .... ...ft( .982\right) ^{20}}}{8}\approx -0.979582 \\ && \\ && \\ && \end{eqnarray*}$$

The exact answers are $x=\displaystyle \frac{7\pm \sqrt{ 209+16\cdot \left( .982\right) ^{20}}}{8}.$ The approximate answers are $x\approx 2.729582$ and -0.979582.

These answers may or may not be the solutions to the original equation. You must check them in the original equation, either by numerical substitution or by graphing.

Numerical Check:

Check the answer $\quad \displaystyle \frac{7+\sqrt{209+16\cdot \left( .982\right) ^{20}}}{8}$ by substituting 2.729582 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.

• Left Side: $\qquad \log _{.982}\left( 4x^{2}-7x-10\right)$ $$\begin{eqnarray*}&& \\ &=&\log _{.982}\left( 4\left( 2.729582\right) ^{2}-7\lef... ...ight) }{\log \left( .982\right) } \\ &\approx &20 \\ && \\ && \end{eqnarray*}$$

• Right Side:$\qquad 20$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 2.729582 for x, then x=2.729582 is a solution.

Check the answer $\quad x=\displaystyle \frac{7-\sqrt{209+16\cdot \left( .982\right) ^{20}}}{8}-0.979582$ by substituting -0.979582 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.

• Left Side: $\qquad \log _{.982}\left( 4x^{2}-7x-10\right)$ $$\begin{eqnarray*}&& \\ &=&\log _{.982}\left( 4\left( -0.979582\right) ^{2}-7\le... ...{\log \left( .982\right) } \\ && \\ &\approx &20 \\ && \\ && \end{eqnarray*}$$

• Right Side:$\qquad 20$

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -0.979582 for x, then x=-0.979582 is a solution.

Graphical Check:

You can also check your answer by graphing

$\quad f(x)=\log _{.982}\left( 4x^{2}-7x-10\right) -20\quad$ (formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis at 2.729582 and -0.979582. This means that 2.729582 and -0.979582 are the real solutions.

If you have trouble graphing the above problem, you might try graphing the equivalent function

$$\begin{eqnarray*}&& \\ f(x) &=&\displaystyle \frac{\log \left( 4x^{2}-7x-10\right) }{\log \left( .982\right) }-20 \\ && \\ && \end{eqnarray*}$$

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